Evaluate $~~\int\ln x\,dx\,$. Choose 1 answer: Choose 1 answer: (Choice A) A $=x\ln(x)-\dfrac1x+C$ (Choice B) B $=x\ln(x)-x+C$ (Choice C) C $=\ln(x)-x+C$ (Choice D) D $=\ln(x)-\dfrac1x+C$
Answer: We will solve this by integrating by parts. We know that $ \int u(x)v\,^\prime(x)dx = u(x)v(x)-\int u\,^\prime(x)v(x)dx\,$. We can rewrite this as $ \int u\ dv = uv-\int v\ du\,$. In this problem we will let $~u = \ln x~$ and $~dv= dx~\,$. Then $~du = \dfrac1xdx~$ and $~v = x\,$. Integration by parts gives $ \int \ln x\,dx = x\ln x-\int x\cdot\dfrac1x\,dx$ $ =x\ln x-x+C\,$.